290 ELECTRICAL APPARATUS

hence the current:

I = - {[e0 cos a — e cos (a + /3 + 5)]

z

— j[e<> sin a — e sin (a + fi + <5)]}

= Jo + y sin I (sin (a + ft + |) + j cos(a + 0 + |) [ (3)

the power:

/a
P = - {e0 COS (a — |3 — 6) — e cos a}

2!

D , 2eeQ . 5 . / 0 S\ ,A.

= p0 + __ sm ^ sm ^ _ 0 _ ^-j . (4)

Let now:

#o = mean velocity (linear, at radius of gyration) of syn-
chronous machine;

s = slip, or decrease of velocity, as fraction of VQ, where s is
a (periodic) function of time; hence

v = VQ (1 — s) = actual velocity, at time, t.

During the time element, dt, the position of the synchronous
motor armature regarding the impressed e.m.f., eo, and thereby
the phase angle, ft + d, of e, changes by:

where:
and
Let:

(5)

0 = 2 vft,

f = frequency of impressed e.m.f., e0.

m = mass of revolving machine elements, and

M o = J^j Wo2 = mean mechanical momentum, reduced to

joules or watt-seconds; then the momentum at time, t, and

velocity v = VQ (1 — s) is:

M = M ^o2 (1 - s)2,
and the change of momentum during the time element, dt, is:

dM
_-

cfe